Factoring Trinomials: Applying the formula

$$The equation in the problem is:

$x^2+6x+9=0$We want to solve this equation using factoring,

First, we'll check if we can factor out a common factor, but this is not possible, since there is no common factor for all three terms on the left side of the equation, we can identify that we can factor the expression on the left side using the quadratic formula for a trinomial squared, however, we prefer to factor it using the factoring method according to trinomials, let's refer to the search for Factoring by trinomials:

Let's note that the coefficient of the squared term (the term with the second power) is 1, so we can try to perform factoring according to the quick trinomial method: (This factoring is also called "automatic trinomial"),

But before we do this in the problem - let's recall the general rule for factoring by quick trinomial method:

The rule states that for the algebraic quadratic expression of the general form:

$x^2+bx+c$We can find a factorization in the form of a product if we can find two numbers $m,\hspace{4pt}n$such that the following conditions are met (conditions of the quick trinomial method):

$\begin{cases} m\cdot n=c\\ m+n=b \end{cases}$If we can find two such numbers $m,\hspace{4pt}n$then we can factor the general expression mentioned above into the form of a product and present it as:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$which is its factored form (product factors) of the expression,

Let's return now to the equation in the problem that we received in the last stage after arranging it:

$x^2+6x+9=0$Let's note that the coefficients from the general form we mentioned in the rule above:

$x^2+bx+c$are:$\begin{cases} c=9 \\ b=6 \end{cases}$where we didn't forget to consider the coefficient together with its sign,

Let's continue, we want to factor the expression on the left side into factors according to the quick trinomial method, above, so we'll look for a pair of numbers $m,\hspace{4pt}n$ that satisfy:

$\begin{cases} m\cdot n=9\\ m+n=6 \end{cases}$We'll try to identify this pair of numbers through logical thinking and using our knowledge of the multiplication table, we'll start from the multiplication between the two required numbers $m,\hspace{4pt}n$ that is - from the first row of the pair of requirements we mentioned in the last stage:

$m\cdot n=9$We identify that their product needs to give a positive result, and therefore we can conclude that their signs are identical,

Next, we'll consider the factors (integers) of the number 9, and from our knowledge of the multiplication table we can know that there are only two possibilities for such factors: 3 and 3, or 9 and 1, as we previously concluded that their signs must be identical, a quick check of the two possibilities regarding the fulfillment of the second condition:

$m+n=6$ will lead to a quick conclusion that the only possibility for fulfilling both of the above conditions together is:

$3,\hspace{4pt}3$That is - for:

$m=3,\hspace{4pt}n=3$(It doesn't matter which one we call m and which one we call n)

It is satisfied that:

$\begin{cases} \underline{3}\cdot \underline{3}=9\\ \underline{3}+\underline{3}=6 \end{cases}$ From here - we understood what the numbers we are looking for are and therefore we can factor the expression on the left side of the equation in question and present it as a product:

$x^2+6x+9 \\ \downarrow\\ (x+3)(x+3)$

In other words, we performed:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

If so, we have factored the quadratic expression on the left side of the equation into factors using the quick trinomial method, and the equation is:

$x^2+6x+9=0 \\ \downarrow\\ (x+3)(x+3)=0\\ (x+3)^2=0\\$$$$$Where in the last stage we noticed that in the expression on the left side the term:

$(x+3)$

multiplies itself and therefore the expression can be written as a squared term:

$(x+3)^2$

Now that the expression on the left side has been factored into a product form (in this case not just a product but actually a power form) we will continue to quickly solve the equation we received:

$(x+3)^2=0$

Let's pay attention to a simple fact, on the left side there is a term raised to the second power, and on the right side the number 0,

and only 0 squared (to the second power) will give the result 0, so we get that the equation equivalent to this equation is the equation:

$x+3=0$(In the same way we could have operated algebraically in a pure form and taken the square root of both sides of the equation, we'll discuss this in a note at the end)

We'll solve this equation by moving the free number to the other side and we'll get that the only solution is:

$x=-3$Let's summarize then the stages of solving the quadratic equation using the quick trinomial factoring method, we got that:

$x^2+6x+9=0 \\ \downarrow\\ (x+3)(x+3)=0\\ (x+3)^2=0\\ \downarrow\\ x+3=0\\ x=-3$Therefore, the correct answer is answer C.

Note:

We could have reached the final equation by taking the square root of both sides of the equation, however - taking a square root involves considering two possibilities: positive and negative (it's enough to consider this only on one side, as described in the calculation below), that is, we could have performed:

$(x+3)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{8pt}}\\ \downarrow\\ \sqrt{(x+3)^2}=\pm\sqrt{0} \\ x+3=\pm0\\ x+3=0$Where on the left side the root (which is a half power) and the second power canceled each other out (this follows from the law of powers for power over power), and on the right side the root of 0 is 0, and we considered two possibilities positive and negative (this is the plus-minus sign indicated) except that the sign (which is actually multiplication by one or minus one) does not affect 0 which remains 0 in both cases, and therefore we reached the same equation we reached with logical and unambiguous thinking earlier - in the solution above,

In any other case where on the right side was a number different from 0, we could have solved only by taking the root etc. and considering the two positive and negative possibilities which would then give two different possibilities for the solution.

Let's solve the given equation:

$x^2-3x+2=0$

Note that the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

We will look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=2\\ m+n=-3\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we are looking for must yield a positive result, therefore we can conclude that both numbers have the same sign, according to multiplication rules, and now we'll remember that the possible factors of 2 are 2 and 1, fulfilling the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are identical will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-2\\ n=-1 \end{cases}$

Therefore we will factor the expression on the left side of the equation to:

$x^2-3x+2=0 \\ \downarrow\\ (x-2)(x-1)=0$

Remember that the product of expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll obtain two simple equations and solve them by isolating the unknown on one side:

$x-2=0\\ \boxed{x=2}$

or:

$x-1=0\\ \boxed{x=1}$

Let's summarize then the solution of the equation:

$x^2-3x+2=0 \\ \downarrow\\ (x-2)(x-1)=0 \\ \downarrow\\ x-2=0\rightarrow\boxed{x=2}\\ x-1=0\rightarrow\boxed{x=1}\\ \downarrow\\ \boxed{x=2,1}$

Therefore the correct answer is answer B.

Let's solve the given equation:

$x^2-12x+36=0$

Note that we can factor the expression on the left side using the perfect square binomial formula:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

We'll do this using the fact that:

$36=6^2$Therefore, we'll represent the rightmost term as a squared term:

$x^2-12x+36=0 \\ \downarrow\\ \textcolor{red}{x}^2-24x+\textcolor{blue}{6}^2=0$

Now let's examine again the perfect square binomial formula mentioned earlier:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side of the equation that we obtained in the last step:

$\textcolor{red}{x}^2-\underline{12x}+\textcolor{blue}{6}^2=0$

Note that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{6}^2$indeed match the form of the first and third terms in the perfect square binomial formula (which are highlighted in red and blue),

However, in order to factor this expression (on the left side of the equation) using the perfect square binomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we'll ask if we can represent the expression on the left side of the equation as:

$\textcolor{red}{x}^2-\underline{12x}+\textcolor{blue}{6}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}}+\textcolor{blue}{12}^2=0$

And indeed it is true that:

$2\cdot x\cdot6=12x$

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

$\textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}}+\textcolor{blue}{6}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{6})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

$(x-6)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-6=\pm0\\ x-6=0\\ \boxed{x=6}$

Let's summarize the solution of the equation:

$x^2-12x+36=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}+\textcolor{blue}{6}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{6})^2=0 \\ \downarrow\\ x-6=0\\ \downarrow\\ \boxed{x=6}$

Therefore the correct answer is answer A.

Let's notice that this is a quadratic equation, therefore we'll first arrange it and move all terms to one side and 0 to the other side:

$x^2=10-9x \\ x^2+9x-10=0$

remembering that when moving a term to the other side its sign changes,

We now want to solve this equation using factoring,

First we'll check if we can factor out a common factor, but this is not possible, since there is no factor common to all three terms on the left side of the equation, therefore, we'll look for trinomial factoring:

Note that the coefficient of the quadratic term (the term with power of 2) is 1, so we can try to use quick trinomial method: (this factoring is also called "automatic trinomial"),

But before we do this in our problem - let's recall the rule for quick trinomial method:

The rule states that for an algebraic quadratic expression in the general form:

$x^2+bx+c$

we can find a factored form if we can find two numbers $m,\hspace{4pt}n$that satisfy the following conditions (quick trinomial method conditions):

$\begin{cases} m\cdot n=c\\ m+n=b \end{cases}$

If we can find two such numbers $m,\hspace{4pt}n$then we can factor the general expression above into a product form and write it as:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

which is its factored form (multiplication factors) of the expression,

Let's return now to the equation in our problem after arranging it in the last step:

$x^2+9x-10=0$

Note that the coefficients from the general form we mentioned in the rule above:

$x^2+bx+c$

are:$\begin{cases} c=-10\\ b=9 \end{cases}$

where we didn't forget to consider the coefficient together with its sign,

Let's continue, we want to factor the expression on the left side using quick trinomial method, above, so we'll look for a pair of numbers $m,\hspace{4pt}n$that satisfy:

$\begin{cases} m\cdot n=-10\\ m+n=9 \end{cases}$

Let's try to find these numbers through logical thinking and using our knowledge of multiplication tables, starting with the multiplication of the two required numbers $m,\hspace{4pt}n$meaning - from the first row of requirements we specified in the last step:

$m\cdot n=-10$

We identify that their product must give a negative result, therefore we can conclude that their signs must be different,

Next we'll consider the factors (whole numbers) of 10, and from our knowledge of multiplication tables we know there are only two possibilities for such factors: 2 and 5, or 10 and 1, since we previously concluded their signs must be different, a quick check of both possibilities regarding the second condition:

$m+n=9$

will lead to the quick conclusion that the only possibility for satisfying both conditions above together is:

$10,\hspace{4pt}-1$

meaning - for:

$m=10,\hspace{4pt}n=-1$

(it doesn't matter which we call m and which we call n)

It satisfies that:

$\begin{cases} \underline{10}\cdot (\underline{-1})=-10\\ \underline{10}+(\underline{-1})=9 \end{cases}$

From here - we understood what numbers we're looking for and therefore we can factor the expression on the left side of the equation in question and present it as a product:

$x^2+9x-10 \\ \downarrow\\ (x+10)(x+(-1))$

Meaning we performed:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

Therefore we factored the quadratic expression on the left side of the equation using quick trinomial method, and the equation is:

$x^2+9x-10=0 \\ \downarrow\\ (x+10)(x+(-1))=0\\ (x+10)(x-1)=0\\$where in the last step we just simplified the expression in the right parentheses in the product,

Now that the expression on the left side is factored we'll continue to its quick solution,

Let's note a simple fact, on the left side there's a product of two terms, and on the right side is 0,

Therefore we can conclude that the only two possibilities for which this equation will be satisfied are if:

$x+10=0$

or if:

$x-1=0$

since only multiplying a number by 0 will give the result -0,

From here we'll solve the two new equations we got:

$x+10=0 \rightarrow x=-10\\ x-1=0\rightarrow x=1$

where we solved each equation separately,

Let's summarize: We therefore got the solutions to the quadratic equation and used quick trinomial method to factor the quadratic expression on its left side:

$x^2+9x-10=0 \\ \downarrow\\ (x+10)(x+(-1))=0\\ (x+10)(x-1)=0\\$which are:

$x_1=-10, \hspace{4pt}x_2=1$

where substituting either of these solutions, the first or the second, in the equation - will yield a true statement,

Therefore the correct answer is answer D.

Let's solve the given equation:

$5x-14=-x^2$

First, let's arrange the equation by moving terms:

$5x-14=-x^2 \\ x^2+5x-14=0$

Now we notice that the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-14\\ m+n=5\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to yield a negative result, therefore we can conclude that the two numbers have different signs, according to multiplication rules, and now we'll remember that the possible factors of 14 are 2 and 7 or 14 and 1, fulfilling the second requirement mentioned, along with the fact that the signs of the numbers we're looking for are different from each other will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=7\\ n=-2 \end{cases}$

Therefore we'll factor the expression on the left side of the equation to:

$x^2+5x-14=0 \\ \downarrow\\ (x+7)(x-2)=0$

From here we'll remember that the result of multiplication between expressions will yield 0 only if at least one of the multiplying expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the unknown in each:

$x+7=0\\ \boxed{x=-7}$

or:

$x-2=0\\ \boxed{x=2}$

Let's summarize the solution of the equation:

$5x-14=-x^2 \\ x^2+5x-14=0 \\ \downarrow\\ (x+7)(x-2)=0 \\ \downarrow\\ x+7=0\rightarrow\boxed{x=-7}\\ x-2=0\rightarrow\boxed{x=2}\\ \downarrow\\ \boxed{x=-7,2}$

Therefore the correct answer is answer C.

Let's notice that this is a quadratic equation, therefore we'll first arrange it and move all terms to one side and have 0 on the other side:

$x^2+3=-4x \\ x^2+4x+3=0$

remembering that when moving a term to the other side its sign changes,

Now we want to solve this equation using factoring,

First we'll check if we can factor out a common factor, but this isn't possible, since there is no factor common to all three terms on the left side of the equation, so we'll look for trinomial factoring:

Note that the coefficient of the quadratic term (the term with power of 2) is 1, therefore we can try to use quick trinomial method: (this factoring is also called "automatic trinomial"),

But before we do this in our problem - let's recall the rule for quick trinomial method:

The rule states that for the quadratic algebraic expression in the general form:

$x^2+bx+c$

we can find a factored form if we can find two numbers $m,\hspace{4pt}n$such that the following conditions (quick trinomial method conditions) are met:

$\begin{cases} m\cdot n=c\\ m+n=b \end{cases}$

If we can find such two numbers $m,\hspace{4pt}n$then we can factor the general expression above into a product form and present it as:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

which is its factored form (multiplication factors) of the expression,

Let's return now to the equation in our problem that we got in the last step after arranging it:

$x^2+4x+3=0$

Note that the coefficients from the general form we mentioned in the rule above:

$x^2+bx+c$ are:$\begin{cases} c=3\\ b=4 \end{cases}$where we didn't forget to consider the coefficient together with its sign,

Let's continue, we want to factor the expression on the left side using quick trinomial method, above, so we'll look for a pair of numbers $m,\hspace{4pt}n$

that satisfy:

$\begin{cases} m\cdot n=3 \\ m+n=4 \end{cases}$

We'll try to find these numbers through logical thinking and using our knowledge of multiplication tables, starting with the multiplication of the two required numbers $m,\hspace{4pt}n$meaning - from the first row of requirements we mentioned in the last step:

$m\cdot n=3$

We can identify that their product needs to give a positive result, therefore we can conclude that their signs must be the same,

Next we'll consider the factors (whole numbers) of 3, and from our knowledge of multiplication tables we can know that there's only one possibility for such factors: 3 and 1, where we previously concluded that their signs must be identical, a quick check of this possibility regarding the second condition (and there's always an obligation to check the condition, even if there's only one possible pair of whole number factors as mentioned above):

$m+n=4$

will lead to the quick conclusion that the only possibility for both conditions above to be met together is:

$3,\hspace{4pt}1$meaning - for:

$m=3,\hspace{4pt}n=1$

(it doesn't matter which we call m and which we call n)

It holds that:

$\begin{cases} \underline{3}\cdot \underline{1}=3\\ \underline{3}+\underline{1}=4 \end{cases}$

From here - we understood what numbers we're looking for and therefore we can factor the expression on the left side of the equation in question and present it as a product:

$x^2+4x+3\\ \downarrow\\ (x+3)(x+1)$

Meaning we performed:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

Therefore we factored the quadratic expression on the left side of the equation using quick trinomial method, and the equation is:

$x^2+4x+3=0 \\ \downarrow\\ (x+3)(x+1)=0\\$

Now that the expression on the left side is factored we'll continue to its quick solution,

Let's note a simple fact, on the left side there's a multiplication between two terms, and on the right side is 0,

Therefore we can conclude that the only two possibilities for which this equation will hold are if:

$x+3=0$

or if:

$x+1=0$

since only multiplying a number by 0 will give the result 0,

From here we'll solve the two new equations we got:

$x+3=0 \rightarrow x=-3\\ x+1=0\rightarrow x=-1$

where we solved each equation separately,

To summarize: We therefore got the solutions to the quadratic equation and used quick trinomial method to factor the quadratic expression on its left side:

$x^2+4x+3=0 \\ \downarrow\\ (x+3)(x+1)=0\\$which are:

$x_1=-3, \hspace{4pt}x_2=-1$

where substituting either of these solutions, the first or the second, in the equation - will yield a true statement,

Therefore the correct answer is answer B.

Let's solve the given equation:

$x^2+3x=10$

First, let's arrange the equation by moving terms:

$x^2+3x=10 \\ x^2+3x-10=0$

Now we notice that the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-10\\ m+n=3\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to yield a negative result, therefore we can conclude that the two numbers have different signs, according to multiplication rules, and now we'll remember that the possible factors of 10 are 2 and 5 or 10 and 1, fulfilling the second requirement mentioned, along with the fact that the numbers we're looking for have different signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=5\\ n=-2 \end{cases}$

Therefore we'll factor the expression on the left side of the equation to:

$x^2+3x-10=0 \\ \downarrow\\ (x+5)(x-2)=0$

From here we'll remember that the result of multiplication between expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll get two simple equations and solve them by isolating the unknown in each:

$x+5=0\\ \boxed{x=-5}$

or:

$x-2=0\\ \boxed{x=2}$

Let's summarize then the solution of the equation:

$x^2+3x=10 \\ x^2+3x-10=0 \\ \downarrow\\ (x+5)(x-2)=0 \\ \downarrow\\ x+5=0\rightarrow\boxed{x=-5}\\ x-2=0\rightarrow\boxed{x=2}\\ \downarrow\\ \boxed{x=-5,2}$

Therefore the correct answer is answer A.

Let's notice that this is a quadratic equation, therefore we'll first arrange it and move all terms to one side and have 0 on the other side:

$3x^2+10x+17=2x^2-3+x \\ 3x^2-2x^2+10x-x+17+3=0\\ x^2+9x+20=0$

where in the first stage we moved all terms to one side while remembering that when moving a term its sign changes and in the second stage we combined like terms,

We now want to solve this equation using factoring,

First we'll check if we can factor out a common factor, but this isn't possible, since there is no factor common to all three terms on the left side of the equation, therefore, we'll look for trinomial factoring:

Note that the coefficient of the squared term (the term with power of 2) is 1, therefore we can try to use quick trinomial factoring: (this factoring is also called "automatic trinomial"),

But before we do this in our problem - let's recall the rule for quick trinomial factoring:

The rule states that for the algebraic quadratic expression in the general form:

$x^2+bx+c$

we can find a factored form if we can find two numbers $m,\hspace{4pt}n$that satisfy the following conditions (quick trinomial method conditions):

$\begin{cases} m\cdot n=c\\ m+n=b \end{cases}$

If we can find two such numbers $m,\hspace{4pt}n$then we can factor the general expression above into a product form and write it as:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

which is its factored form (multiplication factors) of the expression,

Let's return now to the equation we got in the last stage after arranging it:

$x^2+9x+20=0$

Note that the coefficients from the general form we mentioned in the rule above:

$x^2+bx+c$ are:$\begin{cases}c=20 \\ b=9\end{cases}$where we didn't forget to consider the coefficient together with its sign,

Let's continue, we want to factor the expression on the left side using quick trinomial factoring mentioned above, so we'll look for two numbers $m,\hspace{4pt}n$that satisfy:

$\begin{cases}m\cdot n=20 \\ m+n=9\end{cases}$

Let's try to find these numbers through logical thinking and using our multiplication table knowledge, starting with the multiplication of the two required numbers $m,\hspace{4pt}n$meaning - from the first row of requirements we mentioned in the last stage:

$m\cdot n=20$

We can identify that their product must give a positive result, therefore we can conclude that their signs must be the same,

Next we'll look at the factors (whole numbers) of 20, and from our multiplication table knowledge we know there are three possibilities for such factors: 2 and 10, 4 and 5, 20 and 1, where we previously concluded their signs must be identical, a quick check of these possibilities regarding the second condition (and there's always an obligation to check the condition, even if there's only one possible pair of whole number factors as mentioned above):

$m+n=9$

will lead to the quick conclusion that the only possibility for both conditions to be satisfied together is:

$4,\hspace{4pt}5$

meaning - for:

$m=4,\hspace{4pt}n=5$

(it doesn't matter which we call m and which we call n)

It satisfies that:

$\begin{cases} \underline{4}\cdot\underline{5}=20 \\ \underline{4}+\underline{5}=9 \end{cases}$

From here - we understood what numbers we're looking for and therefore we can factor the expression on the left side of the equation in question and present it as a product:

$x^2+9x+20\\ \downarrow\\ (x+4)(x+5)$

Meaning we performed:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

Therefore we factored the quadratic expression on the left side of the equation using quick trinomial factoring, and the equation is:

$x^2+9x+20=0 \\ \downarrow\\ (x+4)(x+5)=0\\$

Now that the expression on the left side is factored we'll continue to its quick solution,

Let's note a simple fact, on the left side there's a multiplication between two terms, and on the right side is 0,

Therefore we can conclude that the only two possibilities for this equation to be satisfied are if:

$x+4=0$

or if:

$x+5=0$

since only multiplying a number by 0 gives the result 0,

From here we'll solve the two new equations we got:

$x+4=0 \rightarrow x=-4\\ x+5=0\rightarrow x=-5$

where we solved each equation separately,

Let's summarize: We therefore got the solutions to the quadratic equation and used quick trinomial factoring to factor the quadratic expression on its left side:

$x^2+9x+20=0 \\ \downarrow\\ (x+4)(x+5)=0\\$which are:

$x_1=-4,\hspace{4pt}x_2=-5$

where substituting either of these solutions, the first or the second, in the equation - will give a true statement,

Therefore the correct answer is answer B.

Let's notice that this is a quadratic equation, therefore we will first arrange it and move all terms to one side and 0 to the other side:

$x^2-3x-5=10+3x+12 \\ x^2-3x-3x-5-10-12=0\\ x^2-6x-27=0$

where in the first stage we moved all terms to one side while remembering that when moving a term its sign changes and in the second stage we combined like terms,

Now we want to solve this equation using factoring,

First we'll check if we can factor out a common factor, but this is not possible, since there is no factor that is common to all three terms on the left side of the equation, therefore, we'll look for trinomial factoring:

Let's note that the coefficient of the quadratic term (the term with power of 2) is 1, therefore we can try to use quick trinomial method: (this factoring is also called "automatic trinomial"),

But before we do this in our problem - let's recall the rule for quick trinomial method:

The rule states that for the quadratic algebraic expression in the general form:

$x^2+bx+c$

we can find a factored form if we can find two numbers $m,\hspace{4pt}n$that satisfy the following conditions (quick trinomial method conditions):

$\begin{cases} m\cdot n=c\\ m+n=b \end{cases}$

If we can find two such numbers $m,\hspace{4pt}n$then we can factor the general expression above into a product form and present it as:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

which is its factored form (multiplication factors) of the expression,

Let's return now to the equation we got in the last stage after arranging it:

$x^2-6x-27=0$

Let's note that the coefficients from the general form we mentioned in the rule above:

$x^2+bx+c$are:$\begin{cases} c=-27 \\ b=-6 \end{cases}$

where we didn't forget to consider the coefficient together with its sign,

Let's continue, we want to factor the expression on the left side using quick trinomial method above, so we'll look for two numbers $m,\hspace{4pt}n$that satisfy:

$\begin{cases} m\cdot n=-27 \\ m+n=-6 \end{cases}$

Let's try to find these numbers through logical thinking and using our multiplication table knowledge, starting with the multiplication of the two required numbers $m,\hspace{4pt}n$meaning - from the first row of requirements we mentioned in the last stage:

$m\cdot n=-27$

We can identify that their product must give a negative result, therefore we can conclude that their signs are different,

Next we'll look at the factors (whole numbers) of 27, and from our multiplication table knowledge we know there are two possibilities for such factors: 3 and 9, 27 and 1, where we previously concluded their signs must be different, a quick check of these possibilities regarding the second condition (and there's always an obligation to check condition satisfaction, even if there's only one possible pair of whole number factors as mentioned above):

$m+n=-6$

will lead to the quick conclusion that the only possibility for satisfying both conditions above together is:

$-9,\hspace{4pt}3$meaning - for:

$m=-9,\hspace{4pt}n=3$

(it doesn't matter which we call m and which we call n)

It satisfies that:

$\begin{cases} (\underline{-9})\cdot\underline{3}=-27 \\ (\underline{-9})+\underline{3}=-6 \end{cases}$

From here - we understood what numbers we're looking for and therefore we can factor the expression on the left side of the equation in question and present it as a product:

$x^2-6x-27\\ \downarrow\\ (x+(-9))(x+3)$

Meaning we performed:

$x^2+bx+c \\ \downarrow\\ (x+m)(x+n)$

Therefore we factored the quadratic expression on the left side of the equation using quick trinomial method, and the equation is:

$x^2-6x-27=0 \\ \downarrow\\ (x+(-9))(x+3)=0\\ (x-9)(x+3)=0\\$where in the second stage we just simplified the expression inside the left parentheses in the above product.

Now that the expression on the left side is factored let's continue to its quick solution,

Let's note a simple fact, on the left side there's a product of two terms, and on the right side the number 0,

Therefore we can conclude that the only two possibilities for which this equation will be satisfied are if:

$x-9=0$

or if:

$x+3=0$

since only multiplying a number by 0 will give the result 0,

From here we'll solve the two new equations we got:

$x-9=0 \rightarrow x=9\\ x+3=0\rightarrow x=-3$

where we solved each equation separately,

Let's summarize: We therefore got the solutions of the quadratic equation and used quick trinomial method to factor the quadratic expression on its left side:

$x^2-6x-27=0 \\ \downarrow\\ (x-9)(x+3)=0\\$which are:

$x_1=9,\hspace{4pt}x_2=-3$

where substituting either of these solutions, the first or the second, in the equation - will give a true statement,

Therefore the correct answer is answer B.